Numerical Analysis: Interpolation and Approximation

Ans 1.
(a)
y= y0+ ( y1− y0 ) x−x0
x1−x0
y=−1+ 1∗x+1
1 y=−1+ x +1=x y=x
(b)
f ( x )=f ( x2 ) + ( x −x2 ) ( f ( x3 )−f ( x1 ) )
2∗∆ x + ( x−x2 )2
[ f ( x1 )−2∗f ( x2 ) + f ( x3 ) ]
2∗∆ x2
f ( x )=0+ ( x −0 ) ( 1+ 1 )
2 + x2 [−1−0+1 ]
2 f ( x ) =x
(c)
Let ,f ( x )=a∗x3 +b∗x2 +c∗x +d So ,−1=−a+b−c +d0=d1=a+b+ c+ d32=8 a+4 b+2 c +d
−1=−a+b−c 1=a+b+ c Adding we get ,0=−a+ b−c+ a+b+ c=2∗b¿ , b=0 So , we get
−1=−a−c32=8∗a+2∗c So ,−8=−8 a−8 c Adding the previous two equations we get ,
32−8=8 a+ 2 c−8 a−8 c24=−6 cc=−4So ,1=a+0−4+0a=5So , we get the equation as
f ( x ) =5∗x3−4∗x
(d)
y= y0+ ( y1− y0 ) x−x0
x1−x0
¿ , y=1+ ( 32−1 ) ( x−1 )
1 y=1+31 x−31 y=31 x−30
(e)
f ( x )=f ( x2 ) + ( x −x2 ) ( f ( x3 )−f ( x1 ) )
2∗∆ x + ( x−x2 )2
[ f ( x1 )−2∗f ( x2 ) + f ( x3 ) ]
2∗∆ x2
f ( x )=1+ ( x−1 ) ( 32−1 )
2 + ( x−1 )2 [ 0−2∗1+32 ]
2 f ( x )=1+ 31∗ ( x−1 )
2 + ( x−1 )2 30
2
f ( x )=1+ 31∗ ( x−1 )
2 +15∗( x−1 )2
Ans 2.
P1 ( x ) =a0 + a1 ( x−x0 ) P1 ( x ) =−2+0.4 ( x−2 ) P1 ( x ) =−2+0.4 x−0.8 P1 ( x ) =−2.8+0.4 x
So , at x=2.7P1 ( x ) =−2.8+1.08=−1.72